Let → b = ^ i + ^ j + λ ^ k , λ ∈ ℝ . If → a is a vector such that...

Let \overset{\to}{a} = x\overset{^}{i} + y\overset{^}{j} + z\overset{^}{k}

So,\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ x & y & z \\ 1 & 1 & \lambda\end{cases}\right| = \overset{^}{i}\left(\lambday - z\right)+ \overset{^}{j}\left(z - \lambdax\right)+ \overset{^}{k}\left(x - y\right)

On comparing with 13\overset{^}{i} - \overset{^}{j} - 4\overset{^}{k} we get,

$⇒ \lambday - z = 13, z - \lambdax = - 1, x - y = - 4$

and $x + y + \lambdaz = - 21$

$⇒$ Clearly, $\lambda = 3, x = - 2, y = 2 and z = - 7</p><p>So,$\overset{\to}{b} - \overset{\to}{a} = 3\overset{^}{i} - \overset{^}{j} + 10\overset{^}{k} and \overset{\to}{b} + \overset{\to}{a} = - \overset{^}{i} + 3\overset{^}{j} - 4\overset{^}{k}$</p><p>$⇒ \left(\overset{\to}{b} - \overset{\to}{a}\right)$\cdot$\left(\overset{^}{k} - \overset{^}{j}\right)+ \left(\overset{\to}{b} + \overset{\to}{a}\right) \cdot \left(\overset{^}{i} - \overset{^}{k}\right) = 11 + 3

$= 14$